A156856 a(n) = 2025*n^2 + n.

2026, 8102, 18228, 32404, 50630, 72906, 99232, 129608, 164034, 202510, 245036, 291612, 342238, 396914, 455640, 518416, 585242, 656118, 731044, 810020, 893046, 980122, 1071248, 1166424, 1265650, 1368926, 1476252, 1587628, 1703054, 1822530

Offset

1,1

Comments

The identity (32805000*n^2 + 16200*n + 1)^2 - (2025*n^2 + n)*(729000*n + 180)^2 = 1 can be written as A157081(n)^2 - a(n)*A156868(n)^2 = 1.

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

From R. J. Mathar, Mar 09 2009: (Start)

a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).

G.f.: 2*x*(1013 + 1012*x)/(1-x)^3. (End)

E.g.f.: x*(2026 + 2025*x)*exp(x). - G. C. Greubel, Jan 28 2022

Mathematica

LinearRecurrence[{3, -3, 1}, {2026, 8102, 18228}, 40]
Table[2025n^2+n, {n, 30}] (* Harvey P. Dale, Sep 12 2024 *)

Prog

(Magma) I:=[2026, 8102, 18228]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];
(PARI) a(n)=n*(2025*n+1) \\ Charles R Greathouse IV, Dec 23 2011
(Sage) [n*(2025*n +1) for n in (1..40)] # G. C. Greubel, Jan 28 2022

Crossrefs

Cf. A156855, A156868, A157081.

A subsequence of A031768.

Sequence in context: A250811 A350869 A031768 * A031543 A031723 A145721

Adjacent sequences: A156853 A156854 A156855 * A156857 A156858 A156859

Keyword

nonn,easy

Author

Vincenzo Librandi, Feb 17 2009

Status

approved